Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $12.5$ years; the standard deviation is $2.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $10.1$ years.
Answer: $12.5$ $10.1$ $14.9$ $7.7$ $17.3$ $5.3$ $19.7$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $12.5$ years. We know the standard deviation is $2.4$ years, so one standard deviation below the mean is $10.1$ years and one standard deviation above the mean is $14.9$ years. Two standard deviations below the mean is $7.7$ years and two standard deviations above the mean is $17.3$ years. Three standard deviations below the mean is $5.3$ years and three standard deviations above the mean is $19.7$ years. We are interested in the probability of a lion living less than $10.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $10.1$ years and the other half $({16\%})$ will live longer than $14.9$ years. The probability of a particular lion living less than $10.1$ years is ${16\%}$.